Balance the reaction for the combustion of pentane: ?C5H12+?O2→?CO2+?H2O Enter the four coefficients in order, separated by commas (e.G., 1,2,3,4), where 1 indicates the absence of a coefficient.

We have 12 hydrogen in right side and we can balance with 6 in the left side. But the number of oxygen is odd. We add 2 in the right side, so we have 24 H, and in the product side we add a 12.

As we add 2 in the C₅H₁₂, we have 10 C, so we must add 10 to the CO₂ in the product side.

Let's count the oxygens: 20 from the CO₂ + 12 from the water = 32.

We add 16 in the reactant side. Balanced equation is:

2C₅H₁₂ + 16O₂→ 10CO₂ + 12H₂O

We also can divide by /2 in order to have the lowest stoichiometry

C₅H₁₂ + 8O₂→ 5CO₂ + 6H₂O

adioabiola adioabiola · Answer 2 · 2021-10-29T16:50:13+02:00

The four coefficients which balance the reaction for the combustion of pentane are; 1, 8, 5, 6.

The reaction involving the combustion of pentane to yield Carbon(IV)oxide, CO2 and Water, H2O can be balanced as follows;

C5H12 + 8O2 → 5CO2 + 6H2O

To check for the authenticity of the coefficients;

There are 5 carbon atoms on the reactants and products side of the chemical equation.

There are 12 hydrogen atoms on the reactants and products side of the chemical equation.

There are 16 oxygen atoms on the reactants and products side of the chemical equation.

Hence, the chemical equation is balanced with coefficients; 1, 8, 5, 6 respectively.